User blog:Mh314159/FOX notation updated
Here is everything so far, up to the structure S1‹S2›(x) where the S's represent strings. I have more structures that go beyond this but don't want to get ahead of myself. I have made some simplifications of the earlier functions and abandoned some ideas that were not precisely defined. I hope this is all valid so far. There are probably more examples than necessary; the actual rules are pretty short. But for me, the examples are part of the fun. for all functions F, Fn(x) = Fn-1x(x) and F0(x) = F(x) f0(x) = f(x) = x+1 f‹0›(x) = fx(x) (ω growth rate) f‹n› and f‹S› are functions and are iterated by powers; n is a natural number and S is a comma-separated string of natural numbers with a nonzero initial term. f‹n›(x) = f‹n-1›f‹n›(x-1)(x) where f‹n›(0) = f‹n-1›(n) f‹S›(x) = f‹T›f‹S›(x-1)(x) where T is S with its first term reduced by 1. f‹S›(0) = f‹T›(s) where s is the final term in S. Zero replacement rules for string where the first one or more terms is zero. Y is any string of terms including possibly empty and Z is a string of n zeroes: f‹Z,a,Y›(x) = f‹N,(a-1),Y›(x) where N is a string of n terms where each term = f‹X,a-1,Y›(x) where X is a string of n x's Example: f‹0,1,2›(3) = f‹ f‹3,0,2›(3) ,0,2›(3) f‹0,0,1,2›(3) = f‹f‹3,3,0,2›(3),f‹3,3,0,2›(3),0,2›(3) f‹0›(1) = f1(1) = 2 f1(2) = f2(2) = 4 f‹0›(2) = f2(2) = f12(2) = f1(4) = 8 f‹0›(3) = f3(3) f‹1›(1) = f‹0›f‹1›(0)(1) = f‹0›2(1) = f‹0›(2) = f2(2) = 8 f‹1›(2) = f‹0›f‹1›(1)(2) therefore 8 iterations f‹1›(3) = f‹0›f‹1›(2)(2) f‹2›(1) = f‹1›f‹2›(0)(1) = f‹1›f‹1›(2)(1) f‹0,1›(2) = f‹f‹2›(2)›(2) f‹1,1›(1) = f‹0,1›f‹1,1›(0)(1) = f‹0,1›f‹0,1›(1)(1) = f‹0,1›f‹f‹1›(1)›(1)(1) f‹1,1›(2) = f‹0,1›f‹1,1›(1)(2) f‹0,2›(1) = f‹f‹1,1›(1),1›(1) Next: replace f with 1, g with 2, etc n and nm and n‹S› are functions n0(x) = n(x) = n-1‹S›(x) where S is a string of n(x-1) x's n(0) = n-1‹n›(n) n‹0›(x) = nx(x) n‹n›(x) = is recursed using the same rule as f‹n›(x), treating n as f. Ex: g(1) = f‹S›(1) where S is a string of 1's with g(0) = f‹2›(2) terms g‹0›(1) = g1(1) = g(1) g1(2) = g(g(2)) = f‹S›(g(2)) where S is a string of length g(g(2)-1) g‹0›(2) = g2(2) = g12(2) = g1(g1(2)) = gp(g1(2)) where p = g1(2) g‹1›(1) = g‹0›g‹1›(0)(1) = g‹0›g‹0›(1)(1) so this iterates g(1) times g‹1›(2) = g‹0›g‹1›(1)(2) g(2) = 2(2) = 1‹S›(2) where S is a string of 2's with 2(1) terms 3‹0›(1) = 31(1) = 3(1) = 2‹S›(1) = where where S is a string of 1's with 3(0) = 2‹3›(3) terms 31(2) = 32(1) = 3(3(1)) 3‹0›(2) = 32(2) = 312(2) = 31(31(2)) = 31(3(3(1))) 3‹1›(1) = 3‹0›3‹0›(1)(1) iterates the 3‹0› function 3‹0›(1) times 3‹0,1›(1) = 3‹3‹1›(1)›(1) 3‹1,1›(1) = 3‹0,1›3‹1,1›(0)(2) = 3‹0,1›3‹0,1›(1)(2) = 3‹0,1›3‹3‹1›(1)›(1)(2) Next: Bracketed string If Sn is a comma-deliminites string of natural number terms with a nonzero first term, then Tn is that string with the first term decreased by one. S1‹0›(x) = [T1‹S1›x(x)]‹S1›x(x) (pulling a strong function into the brackets, growing the function number) Use previous recursion rules treating A as f for all expressions A S1‹S2›(x) = S1‹T2› p(x) = S1‹T2›p-1(S1‹T2›(x)) where p = S1‹S2›(x-1) (iterating the growth of the expression in brackets) S1‹S2›(0) = S1‹T2›(s) where s is the last term in S2 Ex: 1,1‹1,1›(1) = 1,1‹0,1›p(1) = 1,1‹0,1›(1,1‹0,1›(....1)) with 1,1‹1,1›(0) = 1,1‹0,1›(1) iterations. By zero replacement, 1,1‹0,1›(1) = 1,1‹1,1‹1›(1)›(1) and this will eventually recurse to a huge number of iterations of 1,1‹0›(1) where 1,1‹0›(1) = [0,1‹1,1›(1)]‹1,1›(1) = [[[1‹1,1›(1)]‹1,1›(1)]‹1,1›(1)]‹1,1›(1) Category:Blog posts